Question Number 173373 by AgniMath last updated on 10/Jul/22
Answered by mahdipoor last updated on 10/Jul/22
$${p}^{\mathrm{2}} −{rq}={p}^{\mathrm{2}} −\left(−{rp}−{pq}\right)={p}\left({p}+{r}+{q}\right) \\ $$$$\Rightarrow{q}^{\mathrm{2}} −{rp}={q}\left({p}+{r}+{q}\right) \\ $$$$\Rightarrow{r}^{\mathrm{2}} −{pq}={r}\left({p}+{r}+{q}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{p}+{r}+{q}}\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}\right)=\frac{\mathrm{1}}{{p}+{r}+{q}}×\frac{{pq}+{qr}+{rp}}{{pqr}}= \\ $$$$\mathrm{0} \\ $$
Commented by peter frank last updated on 31/Aug/22
$$\mathrm{thanks} \\ $$